- #1
yungman
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Divergence and Curl in cylindrical and spherical co are:
[tex] \nabla \cdot \vec E \;=\; \frac 1 r \frac {\partial r E_r}{\partial r} + \frac 1 r \frac {\partial E_{\phi}}{\partial \phi} + \frac {\partial E_z}{\partial z} \;=\; \frac 1 {R^2} \frac {\partial R^2 E_R}{\partial R} + \frac 1 {R\;sin \;\theta} \frac {\partial sin\;\theta E_{\theta}}{\partial \theta} + \frac 1 {R\;sin \;\theta}\frac {\partial E_{\phi}}{\partial \phi}[/tex]
AND
[tex] \nabla \times \vec A \;=\; \frac 1 r \left|\begin{array}{ccc}\hat r & \hat {\phi} r & \hat z \\\frac {\partial}{\partial r} & \frac {\partial}{\partial \phi} & \frac {\partial}{\partial z} \\ A_r & A_{\phi} & A_z \end{array}\right|\;=\;\frac 1 {R^2 sin \theta} \left | \begin {array}{ccc} \hat R & \hat {\theta} R & \hat {\theta} R\;sin\;\theta \\\frac {\partial}{\partial R} & \frac {\partial}{\partial \theta} & \frac {\partial}{\partial \phi} \\ A_R & R\;A_{\theta} & R\;sin\theta\;A_{\phi} \end{array}\right|[/tex]Both divergence and curl are spatial derivative of a scalar and vector function at a PARTICULAR point in space respectively. My questions are about r and R in the equation:
1) Is r and R the magnitude from origin to the particular point where the divergence and curl is calculate. eg. If I want to calculate the divergence at the point (1,2,3), then [itex]r=R=\sqrt{1^2+2^2+3^2} = \sqrt{14}[/itex]?
2) Also if we perform div and curb of a vector field C at a point pointed by a position vector [tex]\vec P = \left\langle 1,2,3\right\rangle[/tex] then still [itex]\;r=R=\sqrt{1^2+2^2+3^2} = \sqrt{14}\;[/itex] in calculating [itex]\nabla \cdot \vec C \;\hbox { and }\; \nabla \times \vec C\;[/itex].
Thanks for your time.
Alan
[tex] \nabla \cdot \vec E \;=\; \frac 1 r \frac {\partial r E_r}{\partial r} + \frac 1 r \frac {\partial E_{\phi}}{\partial \phi} + \frac {\partial E_z}{\partial z} \;=\; \frac 1 {R^2} \frac {\partial R^2 E_R}{\partial R} + \frac 1 {R\;sin \;\theta} \frac {\partial sin\;\theta E_{\theta}}{\partial \theta} + \frac 1 {R\;sin \;\theta}\frac {\partial E_{\phi}}{\partial \phi}[/tex]
AND
[tex] \nabla \times \vec A \;=\; \frac 1 r \left|\begin{array}{ccc}\hat r & \hat {\phi} r & \hat z \\\frac {\partial}{\partial r} & \frac {\partial}{\partial \phi} & \frac {\partial}{\partial z} \\ A_r & A_{\phi} & A_z \end{array}\right|\;=\;\frac 1 {R^2 sin \theta} \left | \begin {array}{ccc} \hat R & \hat {\theta} R & \hat {\theta} R\;sin\;\theta \\\frac {\partial}{\partial R} & \frac {\partial}{\partial \theta} & \frac {\partial}{\partial \phi} \\ A_R & R\;A_{\theta} & R\;sin\theta\;A_{\phi} \end{array}\right|[/tex]Both divergence and curl are spatial derivative of a scalar and vector function at a PARTICULAR point in space respectively. My questions are about r and R in the equation:
1) Is r and R the magnitude from origin to the particular point where the divergence and curl is calculate. eg. If I want to calculate the divergence at the point (1,2,3), then [itex]r=R=\sqrt{1^2+2^2+3^2} = \sqrt{14}[/itex]?
2) Also if we perform div and curb of a vector field C at a point pointed by a position vector [tex]\vec P = \left\langle 1,2,3\right\rangle[/tex] then still [itex]\;r=R=\sqrt{1^2+2^2+3^2} = \sqrt{14}\;[/itex] in calculating [itex]\nabla \cdot \vec C \;\hbox { and }\; \nabla \times \vec C\;[/itex].
Thanks for your time.
Alan
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